11.5 Chern-Simons forms

11.5.1 Definition

Let $P_j(\mathcal{F})$ be an arbitrary $2j$ -form characteristic class. Since $P_j(\mathcal{F})$ is closed, it can be written locally as an exact form by Poincaré’s lemma. Let us write

$$P_j(\mathcal{F})=\mathrm{d}{Q}_{2j-1}(\mathcal{A},\mathcal{F})\text{\hspace{1em}(11.100)}$$

where $Q_{2j-1}(\mathcal{A},\mathcal{F})\in \mathfrak{g}\otimes {\Omega }^{2j-1}(M)$ . [Warning: This cannot be true globally. If $P_j=\mathrm{d}{Q}_{2j-1}$ globally on a manifold $M$ without boundary, we would have

$${\int }_M{P}_{m/2}={\int }_M\mathrm{d}{Q}_{m-1}={\int }_{\partial M}{Q}_{m-1}=0$$

where $m=\dim M$ .] The $2j-1$ from $Q_{2j-1}(\mathcal{A},\mathcal{F})$ is called the Chern-Simons form of $P_j(\mathcal{F})$ . From the proof of theorem 11.2(b), we find that $Q$ is given by the transgression of $P_j$

$$Q_{2j-1}(\mathcal{A},\mathcal{F})=T{P}_j(\mathcal{A},0)=j{\int }_0^1{\tilde{P}}_j\left(\mathcal{A},{\mathcal{F}}_t,\dots ,{\mathcal{F}}_t\right)dt\text{\hspace{1em}(11.101)}$$

where ${\tilde{P}}_j$ is the polarization of $P_j$ , $\mathcal{F}=\mathrm{d}\mathcal{A}+{\mathcal{A}}^2$ and we set ${\mathcal{A}}^{\prime }={\mathcal{F}}^{\prime }=0$ . Since $Q_{2j-1}$ depends on $\mathcal{F}$ and $\mathcal{A}$ , we explicitly quote the $\mathcal{A}$ -dependence. Of course, ${\mathcal{A}}^{\prime }$ can be put equal to zero only on a local chart over which the bundle is trivial.

Suppose $M$ is an even-dimensional manifold (dim $M=m=2l$ ) such that $\partial M\ne \varnothing$ . Then it follows from Stokes’ theorem that

$${\int }_M{P}_l(\mathcal{F})={\int }_M\mathrm{d}{Q}_{m-1}(\mathcal{A},\mathcal{F})={\int }_{\partial M}{Q}_{m-1}(\mathcal{A},\mathcal{F}).\text{\hspace{1em}(11.102)}$$

The LHS takes its value in integers, and so does the RHS. Thus $Q_{m-1}$ is a characteristic class in its own right and it describes the topology of the boundary $\partial M$ .

11.5.2 The Chern-Simons form of the Chern character

As an example, let us work out the Chern-Simons form of a Chern character ${\mathrm{ch}}_j(\mathcal{F})$ . The connection ${\mathcal{A}}_t$ which interpolates between 0 and $\mathcal{A}$ is

$${\mathcal{A}}_t=t\mathcal{A}\text{\hspace{1em}(11.103)}$$

the corresponding curvature being

$${\mathcal{F}}_t=t\mathrm{d}\mathcal{A}+t^2{\mathcal{A}}^2=t\mathcal{F}+(t^2-t){\mathcal{A}}^2.(11.104)$$

We find from (11.21) that

$$Q_{2j-1}(\mathcal{A},\mathcal{F})=\frac{1}{(j-1)!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^j{\int }_0^1\mathrm{d}t\mathrm{str}(\mathcal{A},{\mathcal{F}}_t^{j-1}).\text{\hspace{1em}(11.105)}$$

For example,

$$Q_1(\mathcal{A},\mathcal{F})=\frac{\mathrm{i}}{2\pi }{\int }_0^1\mathrm{d}t\mathrm{tr}\mathcal{A}=\frac{\mathrm{i}}{2\pi }\mathrm{tr}\mathcal{A}\text{\hspace{1em}(11.106a)}$$

$$\begin{array}{rl}{Q}_3(\mathcal{A},\mathcal{F})& ={\left(\frac{\mathrm{i}}{2\pi }\right)}^2{\int }_0^1\mathrm{d}t\mathrm{str}(\mathcal{A},t\mathrm{d}\mathcal{A}+t^2{\mathcal{A}}^2)\\ & =\frac{1}{2}{\left(\frac{\mathrm{i}}{2\pi }\right)}^2\mathrm{tr}\left(\mathcal{A}\mathrm{d}\mathcal{A}+\frac{2}{3}{\mathcal{A}}^3\right).\end{array}\text{\hspace{1em}(11.106b)}$$

$$\begin{array}{rl}{Q}_5(\mathcal{A},\mathcal{F})& =\frac{1}{2}{\left(\frac{\mathrm{i}}{2\pi }\right)}^3{\int }_0^1\mathrm{d}t\mathrm{str}[\mathcal{A},(t\mathrm{d}\mathcal{A}+t^2{\mathcal{A}}^2{)}^2]\\ & =\frac{1}{6}{\left(\frac{\mathrm{i}}{2\pi }\right)}^3\mathrm{tr}\left[\mathcal{A}(\mathrm{d}\mathcal{A}{)}^2+\frac{3}{2}{\mathcal{A}}^3\mathrm{d}\mathcal{A}+\frac{3}{5}{\mathcal{A}}^5\right].\end{array}\text{\hspace{1em}(11.106c)}$$

Exercise 11.4

Let $\mathcal{F}$ be the field strength of the $\mathrm{SU}(2)$ gauge theory. Write down the component expression of the identity ${\mathrm{ch}}_2(\mathcal{F})=\mathrm{d}{Q}_3(\mathcal{A},\mathcal{F})$ to verify that (cf lemma 10.3)

$$\mathrm{tr}\left[{ϵ}^{\kappa \lambda \mu \nu }{\mathcal{F}}_{\kappa \lambda }{\mathcal{F}}_{\mu \nu }\right]={\partial }_{\kappa }\left[2{ϵ}^{\kappa \lambda \mu \nu }\mathrm{tr}\left({\mathcal{A}}_{\lambda }{\partial }_{\mu }{\mathcal{A}}_{\nu }+\frac{2}{3}{\mathcal{A}}_{\lambda }{\mathcal{A}}_{\mu }{\mathcal{A}}_{\nu }\right)\right].\text{\hspace{1em}(11.107)}$$

11.5.3 Cartan’s homotopy operator and applications

For later purposes, we define Cartan’s homotopy formula following Zumino (1985) and Alvarez-Gaume and Ginsparg (1985). Let

$${\mathcal{A}}_t={\mathcal{A}}_0+t\left({\mathcal{A}}_1-{\mathcal{A}}_0\right){\mathcal{T}}_t=\mathrm{d}{\mathcal{A}}_t+{\mathcal{A}}_t^2\text{\hspace{1em}(11.108)}$$

as before. Define an operator $l_t$ by

$$l_t{\mathcal{A}}_t=0l_t{\mathcal{F}}_t=\delta t\left({\mathcal{A}}_1-{\mathcal{A}}_0\right).\text{\hspace{1em}(11.109)}$$

We require that $l_t$ be an anti-derivative,

$$l_t\left({\eta }_p{\omega }_q\right)=\left(l_t{\eta }_p\right){\omega }_q+(-1{)}^p{\eta }_p\left(l_t{\omega }_q\right)\text{\hspace{1em}(11.110)}$$

for ${\eta }_p\in {\Omega }^p(M)$ and ${\omega }_q\in {\Omega }^q(M)$ . We verify that

$$(\mathrm{d}{l}_t+l_t\mathrm{d}){\mathcal{A}}_t=l_t({\mathcal{F}}_t-{\mathcal{A}}_t^2)=\delta t({\mathcal{A}}_1-{\mathcal{A}}_0)=\delta t\frac{\partial {\mathcal{A}}_t}{\partial t}$$

and

$$\begin{array}{rl}\left(\mathrm{d}{l}_t+l_t\mathrm{d}\right){\mathcal{F}}_t& =\mathrm{d}\left[\delta t\left({\mathcal{A}}_1-{\mathcal{A}}_0\right)\right]+l_t\left[{\mathcal{D}}_t{\mathcal{F}}_t-{\mathcal{A}}_t{\mathcal{F}}_t+{\mathcal{F}}_t{\mathcal{A}}_t\right]\\ & =\delta t[\mathrm{d}({\mathcal{A}}_1-{\mathcal{A}}_0)+{\mathcal{A}}_t({\mathcal{A}}_1-{\mathcal{A}}_0)+({\mathcal{A}}_1-{\mathcal{A}}_0){\mathcal{A}}_t]\\ & =\delta t{\mathcal{D}}_t({\mathcal{A}}_1-{\mathcal{A}}_0)=\delta t\frac{\partial {\mathcal{F}}_t}{\partial t}\end{array}$$

where we have used the Bianchi identity ${\mathcal{D}}_t{\mathcal{F}}_t=0$ . This shows that for any polynomial $S(\mathcal{A},\mathcal{F})$ of $\mathcal{A}$ and $\mathcal{F}$ , we obtain

$$\left(\mathrm{d}{l}_t+l_t\mathrm{d}\right)S\left({\mathcal{A}}_t,{\mathcal{F}}_t\right)=\delta t\frac{\partial }{\partial t}S\left({\mathcal{A}}_t,{\mathcal{F}}_t\right).\text{\hspace{1em}(11.111)}$$

On the RHS, $S$ should be a polynomial of $\mathcal{A}$ and $\mathcal{F}$ only and not of $\mathrm{d}\mathcal{A}$ or $\mathrm{d}\mathcal{F}$ : if $S$ does contain them, $\mathrm{d}\mathcal{A}$ should be replaced by $\mathcal{F}-{\mathcal{A}}^2$ and $\mathrm{d}\mathcal{F}$ by $\mathcal{D}\mathcal{F}-[\mathcal{A},\mathcal{F}]=-[\mathcal{A},\mathcal{F}]$ . Integrating (11.111) over $[0,1]$ , we obtain Cartan’s homotopy formula

$$S\left({\mathcal{A}}_1,{\mathcal{F}}_1\right)-S\left({\mathcal{A}}_0,{\mathcal{F}}_0\right)=\left(\mathrm{d}{k}_{01}+k_{01}\mathrm{d}\right)S\left({\mathcal{A}}_t,{\mathcal{F}}_t\right)\text{\hspace{1em}(11.112)}$$

where the homotopy operator $k_{01}$ is defined by

$$k_{01}S\left({\mathcal{A}}_t,{\mathcal{F}}_t\right)\equiv {\int }_0^1\delta t{l}_{t}S\left({\mathcal{A}}_t,{\mathcal{F}}_t\right).\text{\hspace{1em}(11.113)}$$

To operate $k_{01}$ on $S(\mathcal{A},\mathcal{F})$ , we first replace $\mathcal{A}$ and $\mathcal{F}$ by ${\mathcal{A}}_t$ and ${\mathcal{F}}_t$ , respectively, then operate $l_t$ on $S({\mathcal{A}}_t,{\mathcal{F}}_t)$ and integrate over $t$ .

Example 11.9

Let us compute the Chern-Simons form of the Chern character using the homotopy formula. Let $S(\mathcal{A},\mathcal{F})={\mathrm{ch}}_{j+1}(\mathcal{F})$ and ${\mathcal{A}}_1=\mathcal{A}$ , ${\mathcal{A}}_0=0$ . Since $d{\mathrm{ch}}_{j+1}(\mathcal{F})=0$ , we have

$${\mathrm{ch}}_{j+1}(\mathcal{F})=\left(\mathrm{d}{k}_{01}+k_{01}\mathrm{d}\right){\mathrm{ch}}_{j+1}\left({\mathcal{F}}_t\right)=\mathrm{d}\left[k_{01}{\mathrm{ch}}_{j+1}\left({\mathcal{F}}_t\right)\right].$$

Thus, $k_{01}{\mathrm{ch}}_{j+1}(\mathcal{F})$ is identified with the Chern-Simons form $Q_{2j+1}(\mathcal{A},\mathcal{F})$ . We find that

$$\begin{array}{rl}{k}_{01}{\mathrm{ch}}_{j+1}({\mathcal{F}}_t)& =\frac{1}{(j+1)!}{k}_{01}\mathrm{tr}{\left(\frac{\mathrm{i}\mathcal{F}}{2\pi }\right)}^{j+1}\\ & =\frac{1}{(j+1)!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^{j+1}{\int }_0^1\delta t{l}_t\mathrm{tr}\left({\mathcal{F}}_t^{j+1}\right)\\ & =\frac{1}{j!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^{j+1}{\int }_0^1\delta t\mathrm{str}\left(\mathcal{A},{\mathcal{F}}_t^j\right)\end{array}\text{\hspace{1em}(11.114)}$$

in agreement with (11.105).

Although a characteristic class is gauge invariant, the Chern-Simons form need not be so. As an application of Cartan’s homotopy formula, we compute the change in $Q_{2j+1}(\mathcal{A},\mathcal{F})$ under $\mathcal{A}\to {\mathcal{A}}^g=g^{-1}(\mathcal{A}+\mathrm{d})g$ , $\mathcal{F}\to {\mathcal{F}}^g=g^{-1}\mathcal{F}g$ . Consider the interpolating families ${\mathcal{A}}_t^g$ and ${\mathcal{T}}_t^g$ defined by

$${\mathcal{A}}_t^g\equiv t{g}^{-1}\mathcal{A}g+g^{-1}\mathrm{d}g\text{\hspace{1em}(11.115a)}$$

$${\mathcal{F}}_t^g\equiv \mathrm{d}{\mathcal{A}}_t^g+{\left({\mathcal{A}}_t^g\right)}^2=g^{-1}{\mathcal{F}}_{t}g\text{\hspace{1em}(11.115b)}$$

where ${\mathcal{F}}_t\equiv t\mathcal{F}+(t^2-t){\mathcal{A}}^2$ . Note that ${\mathcal{A}}_0^g=g^{-1}\mathrm{d}g,{\mathcal{A}}_1^g={\mathcal{A}}^g,{\mathcal{F}}_0^g=0$ and ${\mathcal{F}}_1^g={\mathcal{F}}^g$ . Equation (11.112) yields

$$Q_{2j+1}\left({\mathcal{A}}^g,{\mathcal{F}}^g\right)-Q_{2j+1}\left(g^{-1}\mathrm{d}g,0\right)=\left(\mathrm{d}{k}_{01}+k_{01}\mathrm{d}\right)Q_{2j+1}\left({\mathcal{A}}_t^g,{\mathcal{F}}_t^g\right).\text{\hspace{1em}(11.116)}$$

For example, let $Q_{2j+1}$ be the Chern-Simons form of the Chern character ${\mathrm{ch}}_{j+1}(\mathfrak{F})$ . Since $\mathrm{d}{Q}_{2j+1}({\mathcal{A}}_t^g,{\mathcal{F}}_t^g)={\mathrm{ch}}_{j+1}({\mathcal{F}}_t^g)={\mathrm{ch}}_{j+1}({\mathcal{F}}_t)$ , we have

$$\begin{array}{rl}{k}_{01}\mathrm{d}{Q}_{2j+1}\left({\mathcal{A}}_t^g,{\mathcal{F}}_t^g\right)& =k_{01}{\mathrm{ch}}_{j+1}\left({\mathcal{F}}_t^g\right)\\ & =k_{01}{\mathrm{ch}}_{j+1}\left({\mathcal{F}}_t\right)=Q_{2j+1}(\mathcal{A},\mathcal{F})\end{array}\text{\hspace{1em}(11.117)}$$

where the result of example 11.9 has been used to obtain the final equality. Collecting these results, we write (11.116) as

$$Q_{2j+1}\left({\mathcal{A}}^g,{\mathcal{F}}^g\right)-Q_{2j+1}(\mathcal{A},\mathcal{F})=Q_{2j+1}\left(g^{-1}\mathrm{d}g,0\right)+\mathrm{d}{\alpha }_{2j}\text{\hspace{1em}(11.118)}$$

where ${\alpha }_{2j}$ is a $2j$ -form defined by

$$\begin{array}{rl}& {\alpha }_{2j}(\mathcal{A},\mathcal{F},v)\equiv k_{01}{Q}_{2j+1}\left({\mathcal{A}}_t^g,{\mathcal{F}}_t^g\right)\\ & =k_{01}{Q}_{2j+1}\left({\mathcal{A}}_t+v,{\mathcal{F}}_t\right)\end{array}\text{\hspace{1em}(11.119)}$$

where $v\equiv \mathrm{dg}\cdot g^{-1}$ . [Note that $Q_{2j+1}(\mathcal{A},\mathcal{F})=Q_{2j+1}(g\mathcal{A}{g}^{-1},g\mathcal{F}{g}^{-1})$ .] The first term on the RHS of (11.118) is

$$\begin{array}{rl}{Q}_{2j+1}(g^{-1}\mathrm{d}g,0)& =\frac{1}{j!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^{j+1}{\int }_0^1\delta t\mathrm{tr}[g^{-1}\mathrm{d}g\{(t^2-t)(g^{-1}\mathrm{d}g{)}^2{\}}^j]\\ & =\frac{1}{j!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^{j+1}\mathrm{tr}[(g^{-1}\mathrm{d}g{)}^{2j+1}]{\int }_0^1\delta t(t^2-t{)}^j\\ & =(-1{)}^j\frac{j!}{(2j+1)!}{\left(\frac{\mathrm{i}}{2\pi }\right)}^{j+1}\mathrm{tr}[(g^{-1}\mathrm{d}g{)}^{2j+1}](11.120)\end{array}$$

where we have noted that ${\mathcal{F}}_t=(t^2-t)(g^{-1}\mathrm{d}g{)}^2$ and

$${\int }_0^1\delta t(t^2-t{)}^j=(-1{)}^{j}B(j+1,j+1)=(-1{)}^j\frac{(j!{)}^2}{(2j+1)!}$$

$B$ being the beta function. The $2j+1$ form $Q_{2j+1}(g\mathrm{d}g,0)$ is closed and, hence, locally exact: $\mathrm{d}{Q}_{2j+1}(g^{-1}\mathrm{d}g,0)={\mathrm{ch}}_{j+1}(0)=0$ .

As for ${\alpha }_{2j}$ we have, for example,

$$\begin{array}{rl}{\alpha }_2& =\frac{1}{2}{\left(\frac{\mathrm{i}}{2\pi }\right)}^2{\int }_0^1{l}_t\mathrm{tr}[({\mathcal{A}}_t+v){\mathcal{F}}_t-\frac{1}{3}({\mathcal{A}}_t+v{)}^3]\\ & =\frac{1}{2}{\left(\frac{\mathrm{i}}{2\pi }\right)}^2{\int }_0^1\delta t\mathrm{tr}(-t{\mathcal{A}}^2-v\mathcal{A})\\ & =-\frac{1}{2}{\left(\frac{\mathrm{i}}{2\pi }\right)}^2\mathrm{tr}(v\mathcal{A})\end{array}\text{\hspace{1em}(11.121)}$$

where we have noted that

$$\mathrm{tr}{\mathcal{A}}^2=\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }\mathrm{tr}({\mathcal{A}}_{\mu }{\mathcal{A}}_{\nu })=-\mathrm{d}{x}^{\nu }\wedge \mathrm{d}{x}^{\mu }\mathrm{tr}({\mathcal{A}}_{\nu }{\mathcal{A}}_{\mu })=0.$$

Example 11.10

In three-dimensional spacetime, a gauge theory may have a gauge-invariant mass term given by the Chern-Simons three-form (Jackiw and Templeton 1981, Deser et al 1982a, b). Since the Chern-Simons form changes by a locally exact form under a gauge transformation, the action remains invariant. We restrict ourselves to the U(1) gauge theory for simplicity. Consider the Lagrangian (we put $\mathcal{A}=\mathrm{i}A$ , $\mathcal{F}=\mathrm{i}F$ )

$$\mathcal{L}=-\frac{1}{4}{F}_{\mu v}{F}^{\mu v}+\frac{1}{4}m{ϵ}^{\lambda \mu v}{F}_{\lambda \mu }{A}_v\text{\hspace{1em}(11.122)}$$

where $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ . Note that the second term is the Chern-Simons form of the second Chern character $F^2$ (modulo a constant factor) of the U(1) bundle. The field equation is

$${\partial }_{\mu }{F}^{\mu \nu }+m\ast F^{\nu }=0\text{\hspace{1em}(11.123)}$$

where

$$\ast F^{\mu }=\frac{1}{2}{ϵ}^{\mu \kappa \lambda }{F}_{\kappa \lambda }{F}^{\mu \nu }={ϵ}^{\mu \nu \lambda }\ast F_{\lambda }.$$

The Bianchi identity

$${\partial }_{\mu }\ast F^{\mu }=0\text{\hspace{1em}(11.124)}$$

follows from (11.123) as a consequence of the skew symmetry of $F^{\mu \nu }$ . It is easy to verify that the field equation is invariant under a gauge transformation,

$$A_{\mu }\to A_{\mu }+{\partial }_{\mu }\theta \text{\hspace{1em}(11.125)}$$

while the Lagrangian changes by a total derivative,

$$\mathcal{L}\to -\frac{1}{4}{F}^{\mu \nu }{F}_{\mu \nu }+\frac{1}{4}m{ϵ}^{\lambda \mu \nu }{F}_{\lambda \mu }\left(A_{\nu }+{\partial }_{\nu }\theta \right)=\mathcal{L}+\frac{1}{2}m{\partial }_{\nu }\left(\ast F^{\nu }\theta \right).\text{\hspace{1em}(11.126)}$$

Equation (11.106b) shows that the last term on the RHS is identified with

$$Q_3\left(A^{\theta },F^{\theta }\right)-Q_3(A,F)\sim (A+\mathrm{d}\theta )\mathrm{d}A-A\mathrm{d}A\sim \mathrm{d}(\theta \mathrm{d}A).$$

If we assume that $F$ falls off at large spacetime distances, this term does not contribute to the action:

$$\int {\mathrm{d}}^{3}x\mathcal{L}\to \int {\mathrm{d}}^{3}x\mathcal{L}+\frac{m}{2}\int {\mathrm{d}}^{3}x{\partial }_{\nu }(\ast F^{\nu }\theta )=\int {\mathrm{d}}^{3}x\mathcal{L}.\text{\hspace{1em}(11.127)}$$

Let us show that (11.122) describes a massive field. We first write (11.123) as

$${ϵ}^{\mu \nu \alpha }{\partial }_{\mu }\ast F_{\alpha }=-m\ast F^{\nu }.$$

Multiplying ${\epsilon }_{\kappa \lambda \nu }$ on both sides, we have

$${\partial }_{\lambda }\ast F_{\kappa }-{\partial }_{\kappa }\ast F_{\lambda }=-m{F}_{\kappa \lambda }.$$

Taking the ${\partial }^{\lambda }$ -derivative and using (11.124), we find that

$$\left({\partial }^{\lambda }{\partial }_{\lambda }+m^2\right)\ast F_{\kappa }=0\text{\hspace{1em}(11.128)}$$

which shows that $\ast F_{\kappa }$ is a massive vector field of mass $m$ .

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